A 220\ μF capacitor charges through a 10\ k resistor. Find the time constant. [2 marks]

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How does a capacitor charging through a resistor create a time delay?

Capacitors and time delays: charge stored on a capacitor, the RC time constant, and how a charging capacitor in a potential divider produces a time delay.

An Eduqas GCSE Electronics answer on capacitors and time delays: what a capacitor stores, the charge equation Q = CV, the RC time constant, the exponential charge and discharge of a capacitor through a resistor, and how this makes a time-delay circuit.

Generated by Claude Opus 4.813 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

Eduqas wants you to describe what a capacitor stores, use the charge equation $Q = CV$ , calculate the RC time constant, and explain how a capacitor charging through a resistor produces a time delay. Capacitors and the time constant underpin every timing circuit later in the course, including the 555 monostable and astable.

The answer

What a capacitor stores

Charging and discharging through a resistor

The time constant

Making a time delay

Designing a simple time delay

You need a capacitor to reach about $63\%$ of the supply (one time constant) in $2.0\ \text{s}$ , charging through a resistor. If you use a $100\ \mu\text{F}$ capacitor, what resistor is needed?

step 1: Write the time-constant equation

We want $\tau = 2.0\ \text{s}$ and $\tau = RC$ , so $R = \dfrac{\tau}{C}$ .

step 2: Substitute with base units

$C = 100\ \mu\text{F} = 100 \times 10^{-6}\ \text{F} = 1.0 \times 10^{-4}\ \text{F}$ . Then $R = \dfrac{2.0}{1.0 \times 10^{-4}} = 2.0 \times 10^{4}\ \Omega = 20\ \text{k}\Omega$ .

step 3: Check the result

With $R = 20\ \text{k}\Omega$ and $C = 100\ \mu\text{F}$ , $\tau = 20\,000 \times 100 \times 10^{-6} = 2.0\ \text{s}$ , so the capacitor reaches about $63\%$ of the supply after $2.0\ \text{s}$ , as required.

Examples in context

Capacitors and the time constant appear throughout the course. The 555 monostable times a pulse by charging a capacitor through a resistor to two-thirds of the supply; the 555 astable charges and discharges a capacitor repeatedly to make an oscillator. Smaller capacitors smooth power supplies and remove switch bounce. A time-delay circuit (lights that stay on for a set time, a delayed alarm) is a popular non-exam assessment idea built directly on $\tau = RC$ .

Try this

Q1. Write the equation for the charge stored on a capacitor. [1 mark]

Cue. $Q = CV$ .

Q2. A $220\ \mu\text{F}$ capacitor charges through a $10\ \text{k}\Omega$ resistor. Find the time constant. [2 marks]

Cue. $\tau = RC = 10\,000 \times 220 \times 10^{-6} = 2.2\ \text{s}$ .

Q3. State approximately what fraction of the supply voltage a capacitor reaches after one time constant when charging. [1 mark]

Cue. About $63\%$ .

Exam-style practice questions

Practice questions written in the style of WJEC Eduqas exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Eduqas 20203 marksA

100\ \mu\text{F}

capacitor charges through a

47\ \text{k}\Omega

resistor. Calculate the time constant and state, approximately, the percentage of the final voltage the capacitor reaches after one time constant.

Show worked answer →

Time constant: $\tau = RC$ . Convert: $R = 47\ \text{k}\Omega = 47\,000\ \Omega$ and $C = 100\ \mu\text{F} = 100 \times 10^{-6}\ \text{F}$ .

Substitute: $\tau = 47\,000 \times 100 \times 10^{-6} = 4.7\ \text{s}$ .

After one time constant a charging capacitor reaches about $63\%$ of its final (supply) voltage.

Markers reward the conversion, $\tau = 4.7\ \text{s}$ , and the figure $63\%$ . The usual error is leaving the capacitance in microfarads, giving an answer a million times too large.

Eduqas 20223 marksA capacitor is charged to

6.0\ \text{V}

. Calculate the charge stored if its capacitance is

220\ \mu\text{F}

, and explain how increasing the series resistance would change the time taken to charge it.

Show worked answer →

Charge stored: $Q = CV = (220 \times 10^{-6}) \times 6.0 = 1.32 \times 10^{-3}\ \text{C} = 1.32\ \text{mC}$ .

Effect of larger resistance (up to 2 marks): the time constant $\tau = RC$ increases, so the capacitor charges more slowly and takes longer to reach a given voltage. A larger resistor limits the charging current, so the charge builds up more slowly.

Markers reward $Q = 1.32\ \text{mC}$ and a correct explanation that a bigger $R$ gives a bigger $\tau$ and a slower charge.

Related dot points

Sources & how we know this

WJEC Eduqas GCSE (9-1) Electronics specification (C490) — WJEC Eduqas (2017)